Problem: Complete the square to solve for $x$. $x^{2}-6x-27 = 0$
Solution: Begin by moving the constant term to the right side of the equation. $x^2 - 6x = 27$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-6$ , half of it would be $-3$ , and squaring it gives us ${9}$ $x^2 - 6x { + 9} = 27 { + 9}$ We can now rewrite the left side of the equation as a squared term. $( x - 3 )^2 = 36$ Take the square root of both sides. $x - 3 = \pm6$ Isolate $x$ to find the solution(s). $x = 3\pm6$ So the solutions are: $x = 9 \text{ or } x = -3$ We already found the completed square: $( x - 3 )^2 = 36$